Author Topic: Colloidal Silver reduced with Maltose and capped with Gelatine  (Read 9916 times)

Offline kephra

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #15 on: June 18, 2015, 04:25:30 PM »
Maltose is a weaker reducing agent than glucose, and its a heavier molecule which should make it a better stabilizer. 
For substitution, glucose should be half the amount by weight.
Colloidal Silver is only a bargain if you make it yourself.

Offline RickinWI

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #16 on: June 18, 2015, 05:14:11 PM »
Excellent Kephra, thanks for the explanation.
I was looking for a forum glossary, to understand if some of these words, (reactions), had more than one meaning.
I now understand stabilizing and capping as the same thing.
I'm under the impression that stabilizing is important for long term storage, and capping helps to get the silver past the stomach acid.

So, the gelatin process mostly adds some extra "capping" then?
I assume that it's the different molecular structure of each sugar type that determines it's effects?
Fascinating!

-Sancho

I should clarify: In my earlier post I mentioned "stability" of colloidal silver that was reduced with Karo, Dextrose etc at higher pH values. In that context I was referring to the stability of colloidal silver that was just reduced but not purposely capped with something else (Gelatin) afterward.

I guess all of the sugar reducers also have somewhat of a capping effect, however once the colloidal silver hits the stomach acid, any glucose cap is dissolved right off immediately exposing our carefully made silver particles to the ravages of the HCl acid in the stomach. Cinnamon Reduced colloidal silver holds up a little better & longer but utimately fails in a relatively short amount of time if exposed to the most severe conditions (pH 1.0 HCl acid and salt). You can do a similar experiment yourself very easily even if you don't have any HCl acid. Just take some colloidal silver and add some vinegar (pH about 4.0) and or salt. Even salt alone destroys glucose reduced colloidal silver. If salty vinegar can destroy colloidal silver so easily imagine what salty HCl acid in the stomach can do (pH about 2.0---3.5)

What puzzles me after doing these experiments is why I was getting good efficacy before using glucose reduced colloidal silver. Maybe it does not matter if the stomach breaks down the colloidal silver. Obviously all the silver is still there. It is just no longer yellow colloidal silver as we know it. My guess is that once the stomach acid breaks down the colloidal silver it converts it to IS which would then rapidly form silver chloride?  Maybe Kephra could shed some light on this for us.
So many VARIABLES & so little TIME.

Offline kephra

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #17 on: June 18, 2015, 05:20:59 PM »
There was an experiment done with artificial stomach juice which showed that 75% of the silver nanoparticles survived the stomach intact in the amount of time it would have taken for the stomach to empty.

Colloidal Silver is only a bargain if you make it yourself.

Offline RickinWI

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #18 on: June 18, 2015, 05:25:38 PM »
Maltose is a weaker reducing agent than glucose, and its a heavier molecule which should make it a better stabilizer. 
For substitution, glucose should be half the amount by weight.

I have been using about 150 mg/L (0.15gm/L) of Maltose for reducing 20 PPM colloidal silver.  Is that enough? What's the stoichiometric amount for Malt? (80 mg/L ?)  I think I remember that 40 mg/L is the minimum if using Dextrose.
So many VARIABLES & so little TIME.

Offline PeterXXL

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #19 on: June 18, 2015, 07:36:20 PM »
RickinWI: I also regard the Maltose reduced colloidal silver to be clearer than when using Glucose. But a big difference, is that it was possible to make a 500 ppm batch of colloidal silver with Maltose (and Gelatine) withut problem.

So, if you use the exact same procedure but substitute pure Dextrose for Maltose it does not work (as well)?


Confirmed, it does not work (trying to make a 500 ppm colloidal silver using Glucose/Dextrose and Gelatin, as the result was not clear enough). I have not yet tested Maltodextrin though. But using Cinnamon extract also works fine - making the result very clear. With Cinnamon extract, the result is clear enough but much darker and with the typical Cinnamon smell. And the latter is a reason for me to avoid it.

Offline PeterXXL

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #20 on: June 18, 2015, 09:10:02 PM »
Maltose is a weaker reducing agent than glucose, and its a heavier molecule which should make it a better stabilizer. 
For substitution, glucose should be half the amount by weight.

I have been using about 150 mg/L (0.15gm/L) of Maltose for reducing 20 PPM colloidal silver.  Is that enough? What's the stoichiometric amount for Malt? (80 mg/L ?)  I think I remember that 40 mg/L is the minimum if using Dextrose.


I calculated it like this...

I wanted to make a 250 ml of colloidal silver with a concentration of 500 ppm.

250 ml 500 ppm = 0.25 * 500 = 125 mg Silver.

Atom weight / molar mass of Silver = 107.87
Atom weight / molar mass of Maltose = 342.30

125 mg Silver / 107.87 weight of Silver = 1.1588 mol of Silver

Assuming that 1 molecule of Maltose will release 1 electron to reduce 1 Silver ion (looking at the molecule structure of Maltose, it look like it can release more than 1 electrons though)...

1.1588 mol of Silver * 342.30 atom weight of Maltose = 396.658 mg of Maltose is needed for the reduction
396.658 mg = 0.4 gram if we round it up (which should be done in order to be sure that all Silver ions are reduced).

And this was what I also added, beside 0.25 ml of 1-mol sodium carbonate solution and 0.2 gram Gelatin.

Offline wgpeters

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #21 on: June 18, 2015, 09:40:54 PM »
Good answer except that its millimoles of silver (.125 / 107 ) and the ratio is off by two.


Reducing sugars have a carboxyl group at one end of the molecule that does the reduction.  The carboxyl group takes on an oxygen creating a carbon dioxide molecule which is released as gas. 


If you add the sugar to cold ionic silver and then warm it and let it set for a while you will see the CO2 bubbles form on the side of the bottle. 


So 1 molecule of maltose will reduce one molecule of silver oxide releasing two atoms of silver.


Then the stoichiometric amount is half what you calculated.   Better too much than too little though, and I always use more than the stoichiometric amount. 











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Offline RickinWI

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #22 on: June 18, 2015, 10:26:42 PM »
I think I got lost somewhere between the milli-moles & the 500 PPM.

So is the stoichiometric amount of Maltose & Dextrose I mentioned correct for 20 PPM?

Dextrose = 40 mg per Liter (0.04 gm) ?
Maltose = 80 mg per Liter (0.08 gm) ?
So many VARIABLES & so little TIME.

Offline kephra

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #23 on: June 18, 2015, 10:31:16 PM »
Using Peter's example, you should now be able to calculate it.
Tell me your answer, and I will check it.
Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #24 on: June 18, 2015, 10:37:54 PM »
Good answer except that its millimoles of silver (.125 / 107 ) and the ratio is off by two.

Reducing sugars have a carboxyl group at one end of the molecule that does the reduction.  The carboxyl group takes on an oxygen creating a carbon dioxide molecule which is released as gas. 

If you add the sugar to cold ionic silver and then warm it and let it set for a while you will see the CO2 bubbles form on the side of the bottle. 

So 1 molecule of maltose will reduce one molecule of silver oxide releasing two atoms of silver.

Then the stoichiometric amount is half what you calculated.   Better too much than too little though, and I always use more than the stoichiometric amount.


Ok, Yes, it's silver oxide that is reduced and not free silver ions Ag+ so only half that weight of maltose is needed then, rounded up. We have to remember that.

Is the reduction like this then...

Ag2O + C12H22O11 + O => 2 Ag + C11H22O10 + CO2

...or?

Offline wgpeters

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #25 on: June 18, 2015, 10:41:52 PM »
Yes, the formula looks right.
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Offline PeterXXL

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #26 on: June 18, 2015, 10:55:42 PM »
I think I got lost somewhere between the milli-moles & the 500 PPM.

So is the stoichiometric amount of Maltose & Dextrose I mentioned correct for 20 PPM?

Dextrose = 40 mg per Liter (0.04 gm) ?
Maltose = 80 mg per Liter (0.08 gm) ?

Silver atom weight / molar weight = 107.87
Glucose / Dextrose atom weight / molar weight = 180.16
Maltose atom weight / molar weight = 342.30

So...

1 liter 20 ppm reduced with Glucose/Dextrose:
20 mg silver / atom weight of silver 107.87 = 0.1854 milli-mol
0.1854 milli-mol silver * 180.16 atom weight of glucose = 33.40 mg glucose/dextrose
And since 1 molecule of glucose will reduce 1 molecule of silver oxide, which contains 2 silver atoms, only half that weight of glucose/dextrose is needed...
33.40 / 2 = 16.7 mg, which is 0.0167 gram, we round up to 0.02 gram (or more).

1 liter 20 ppm reduced with Maltose:
20 mg silver / atom weight of silver 107.87 = 0.1854 milli-mol
0.1854 milli-mol silver * 342.30 atom weight of maltose = 63.46 mg maltose
And half of that...
63.46 / 2 = 31.73 mg = 0.03173 gram, which is rounded up to 0.04 gram (or more).
« Last Edit: June 18, 2015, 11:44:36 PM by PeterXXL »

Offline RickinWI

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #27 on: June 18, 2015, 11:01:45 PM »
Using Peter's example, you should now be able to calculate it.
Tell me your answer, and I will check it.

Oh shoot, a quiz & I didn't prepare  ???

Ok, my guess for 1 Liter of 20 PPM Maltose Reduced is 0.032 gm = 32 mg?
So many VARIABLES & so little TIME.

Offline RickinWI

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #28 on: June 18, 2015, 11:06:19 PM »
Oh shoot, Peter wrote the right answer on the board before I was done scratching my head   ;)
So many VARIABLES & so little TIME.

Offline PeterXXL

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Re: Colloidal Silver reduced with Maltose and capped with Gelatine
« Reply #29 on: June 19, 2015, 12:06:57 AM »

How to calculate how much of a reducing agent that is needed to reduce silver oxide (ionic silver) made from/during electrolysis.


Atom weights / molar weights
Silver = 107.87
Glucose / Dextrose = 180.16
Maltose = 342.30
Maltodextrin = 3 to 17 times that of Glucose/Dextrose =  540.48 to 3062.17
Note: Maltodextrin contains 3 to 17 glucose molecules, but normally something in the middle, but to be sure we use the highest weight.


PPM to milligram (mg)
First we must convert ppm (concentration in volume) to milligram (concentration in weight).
Weight in milligram = Amount of water in ml / 1000 * ppm.


So a 500 ml of 20 ppm colloidal silver solution contains 500 / 1000 * 20 = 10 mg silver.
Or a 2000 ml (2 liter) 40 ppm colloidal silver solution contains 2000 / 1000 * 40 = 80 mg.


Amount of reducing agent
Then we can calculate how much reducing agent that is needed in milligram, assuming that 1 molecule of reducing agent is able to reduce 1 molecule of silver oxide (which contains 2 silver atoms).


Amount of reducing agent in milligram = ( mg of silver / atom weight of silver )  * atom weight of reducing agent / 2


Examples
So 20 mg of silver requires so much glucose = ( 20 / 107.87 ) * 180.16 / 2 = 16.70 mg.
And 20 mg of silver requires so much maltose = ( 20 / 107.87 ) * 342.30 / 2 = 31.73 mg.
And 20 mg of silver requires so much maltodextrin ( 20 / 107.87 ) * 3062.17 / 2 = 283.88 mg.


Note: 1 mg = 0.001 gram; 10 mg = 0.01 gram; 100 mg = 0.1 gram, etc.